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Functions, inverses and the modulus

A function is a machine with paperwork: a domain of allowed inputs, a range of possible outputs, and rules for chaining machines together or running one backwards. The modulus joins in as the fold-it-positive machine, whose V-shaped graphs make equations sprout extra solutions.

Year 12-13EDEXCEL 9MA0 2.7, 2.8, 2.9

Builds on Graphs, proportion and transformations.

IN THIS TOPIC

  • Work with domain, range, and one-one against many-one mappings.
  • Build composite functions in the right order, and find inverses with their domains.
  • Sketch modulus graphs and solve modulus equations and inequalities.

WHAT YOU PROBABLY THINK

Every function has an inverse.

Machines with paperwork

A function assigns to each input exactly one output, written f(x) or f: x ↦ 2x + 1. Its domain is the set of inputs it accepts; its range is the set of outputs it actually produces. A function is one-one if distinct inputs always give distinct outputs, and many-one if two inputs can share an output, as x2 does with 2 and −2 both landing on 4. The distinction looks bureaucratic and turns out to run the whole lesson.

WORKED EXAMPLE

Composites, in the right order

With f(x) = 2x + 1 and g(x) = x2, find fg(2) and gf(2), and the formula for fg(x).

fg means g first: g(2) = 4, then f(4) = 9, so fg(2) = 9.

The other order: f(2) = 5, then g(5) = 25, so gf(2) = 25.

In general fg(x) = f(x2) = 2x2 + 1.

The results differ, so composition order matters and the notation reads right to left, the function nearest the input acting first.

Inverses, and who gets one

The inverse f−1 undoes f, so f−1f(x) = x for every x in the domain, and its graph is the graph of f reflected in y = x, because undoing swaps input with output. But only one-one functions have inverses, which retires the opening lie. A many-one function sends two inputs to one output, and the would-be inverse could not decide which input to return. The standard rescue is a restricted domain: x2 + 4 on x ≥ 0 keeps only one arm of the parabola and becomes invertible.

The function x squared plus 4 on the domain x at least 0, and its inverse the square root of x minus 4, drawn at equal scales as reflections of each other in the dashed line y equals xf: x² + 4, x ≥ 0f⁻¹: √(x − 4)y = xdomain and range swap with the axes
FIG. 1f(x) = x² + 4 on x ≥ 0 and its inverse √(x − 4), reflections in y = x. The domain restriction is what makes the reflection a function at all.

WORKED EXAMPLE

An inverse with its paperwork

Find the inverse of f(x) = x2 + 4 with domain x ≥ 0, stating the domain and range of f−1.

Write y = x2 + 4 and solve for x: x = √(y − 4), taking the positive root because the domain of f is x ≥ 0.

Swap letters: f−1(x) = √(x − 4).

The range of f is y ≥ 4, so the domain of f−1 is x ≥ 4 and its range is y ≥ 0: domain and range trade places under inversion.

Stating the swapped domain and range is not decoration; without the domain, √(x − 4) is a different function and the answer mark goes with it.

The modulus

The modulus |x| strips signs, so the graph of y = |ax + b| is the line ax + b with everything below the axis folded up, a V with its vertex at the line's root. The same fold applied after a function gives y = |f(x)|; applied inside, y = f(−x) reflects in the y-axis instead. Solving a modulus equation is a two-graph problem, and the sketch does the counting before any algebra starts.

The V-shaped graph of y equals the modulus of 2x minus 1 cut by the line y equals x: they cross at x equal to one third on the reflected arm and at x equal to 1 on the original army = |2x − 1|y = xx = ⅓x = 1one crossing per arm; the sketch counts them first
FIG. 2y = |2x − 1| against y = x. One crossing on the reflected arm at x = ⅓, one on the original arm at x = 1, and the sketch promises exactly two.

WORKED EXAMPLE

A modulus equation, both arms

Solve |2x − 1| = x.

Original arm: 2x − 1 = x gives x = 1, and the check |1| = 1 holds.

Reflected arm: −(2x − 1) = x gives 1 − 2x = x, so x = ⅓, and |−⅓| = ⅓ holds.

Solutions: x = ⅓ and x = 1, matching the sketch's two crossings.

Each arm contributes one candidate, and each candidate is checked in the original equation, because folding can invent solutions that belong to neither arm.

YOUR TURN

The matching inequality

Solve |2x − 1| > x, before opening the working.

Show the working

The critical points are the crossings just found, x = ⅓ and x = 1.

Between them the V dips below the line, so the modulus wins outside: x < ⅓ or x > 1.

A test value settles any doubt: at x = ½ the left side is 0 and the right is ½, so the middle interval genuinely fails.

TRY IT UNSEEN

A composite with a domain trap

With f(x) = 2x + 1 and g(x) = x2, solve gf(x) = 9.

Show the working

gf(x) = (2x + 1)2, so (2x + 1)2 = 9 and 2x + 1 = ±3.

The two arms give x = 1 and −2.

Squares behave like a modulus here, two preimages for one output, which is the many-one story of this lesson, restaged as an exam question.

THE EXAM BIT

  • fg means g first; write the inner function's output before touching the outer one.
  • To invert, set y = f(x), solve for x, swap letters, and state the new domain and range; the domain of f−1 is the range of f.
  • Only one-one functions invert; if asked why f has no inverse, name two inputs sharing an output.
  • Sketch modulus equations first and count crossings; then solve one linear equation per arm and check each answer in the original.
  • For |f(x)| sketches, fold below-axis parts up and leave the rest alone; for f(−x), reflect in the y-axis.

CHECK YOURSELF

With f(x) = 3x − 2 on all of ℝ, find f−1(x), and solve |3x − 2| = 4.

Show a hint

Linear functions are one-one everywhere; the modulus needs both arms.

Show the answer

Solving y = 3x − 2 for x gives x = (y + 2)/3, so f−1(x) = (x + 2)/3, no restriction needed.

The arms of |3x − 2| = 4: 3x − 2 = 4 gives x = 2, and 3x − 2 = −4 gives x = −⅔.

Solutions x = 2 and x = −⅔, and both survive substitution back into the modulus.

fg runs g first; inverses undo, reflect in y = x, and swap domain with range.

Only one-one functions invert; the modulus folds graphs and doubles arms.

CHECK YOUR PROGRESS

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  • Work with domain, range, and one-one against many-one mappings.
  • Build composite functions in the right order, and find inverses with their domains.
  • Sketch modulus graphs and solve modulus equations and inequalities.

No animated video for this topic yet; these notes stand alone.